The Keller-Segel model and the logarithmic Hardy-Littlewood-Sobolev inequality

The Keller-Segel model and the logarithmic Hardy-Littlewood-Sobolev inequality

Jean Dolbeault

http://www.ceremade.dauphine.fr/∼dolbeaul http://www.sciencesmaths-paris.fr/

Ceremade, Universit´e Paris-Dauphine March 18, 2012

Kaust

A – The two-dimensional

parabolic-elliptic Keller-Segel model

Outline

Introduction: the two-dimensional parabolic-elliptic Keller-Segel model, the M<8π regime, scalings, etc

The asymptotic behaviour of the solutions of the Keller-Segel model for small mass

References on the general theory of Keller-Segel systems:

[ Horstmann ]

Disclaimer: many references !

Introduction

The parabolic-elliptic Keller and Segel system







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

∂u

∂t = ∆u− ∇ ·(u∇v) x ∈R², t >0

−∆v =u x ∈R², t >0 u(·,t = 0) =n0≥0 x ∈R² We make the choice:

v(t,x) =− 1 2π

Z

R²

log|x−y|u(t,y)dy and observe that

∇v(t,x) =− 1 2π

Z

R²

x−y

|x−y|²u(t,y)dy Mass conservation: d

dt Z

R²

u(t,x)dx= 0

Blow-up

M =R

R²n0dx>8πandR

R²|x|²n0dx<∞: blow-up in finite time a solution uof

∂u

∂t = ∆u− ∇ ·(u∇v) satisfies

d dt

Z

R²|x|²u(t,x)dx

=− Z

R²

2x· ∇u dx

| {z }

−4M

+ 1 2π

Z Z

R²×R²

2x·(y−x)

|x−y|² u(t,x)u(t,y)dx dy

| {z }

(x−y)·(y−x)

|x−y|² u(t,x)u(t,y)dx dy

= 4M−M²

2π <0 if M>8π

Existence and free energy

M =R

R²n0dx≤8π: global existence [ J¨ager, Luckhaus ],[ JD, Perthame ], [ Blanchet, JD, Perthame ],[ Blanchet, Carrillo, Masmoudi ]

Ifu solves

∂u

∂t =∇ ·[u(∇(logu)− ∇v)]

the free energy

F[u] :=

Z

R²

ulogu dx−1 2 Z

R²

u v dx satisfies

d

dtF[u(t,·)] =− Z

R²

u|∇(logu)− ∇v|²dx

Log HLS inequality[ Carlen, Loss ]: F is bounded from below if M <8π

The dimension d = 2

In dimension d, the normL^d/2(R^d)is critical. Ifd= 2, the mass is critical

Scale invariance: if (u,v)is a solution inR²of the parabolic-elliptic Keller and Segel system, then

λ²u(λ²t, λx), v(λ²t, λx) is also a solution

ForM <8π, the solution vanishes ast → ∞, but saying that

”diffusion dominates” is not correct: to see this, study

”intermediate asymptotics”

The existence setting

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∂u

∂t = ∆u− ∇ ·(u∇v) x ∈R², t >0

−∆v =u x ∈R², t >0 u(·,t = 0) =n0≥0 x ∈R² Initial conditions

n0∈L¹₊(R²,(1+|x|²)dx), n0logn0∈L¹(R²,dx), M:=

Z

R²

n0(x)dx <8π Global existence and mass conservation: M=R

R²u(x,t)dx for any t ≥0, see [ J¨ager-Luckhaus ],[ Blanchet, JD, Perthame ]

v =−2π¹ log| · | ∗u

Time-dependent rescaling

u(x,t) = 1 R²(t)n

x R(t), τ(t)

and v(x,t) =c x

R(t), τ(t)

withR(t) =√

1 + 2t andτ(t) = logR(t)

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∂n

∂t = ∆n− ∇ ·(n(∇c−x)) x∈R², t >0 c=− 1

2π log| · | ∗n x∈R², t >0 n(·,t = 0) =n0≥0 x∈R²

[ Blanchet, JD, Perthame ] Convergence in self-similar variables

tlim→∞kn(·,·+t)−n_∞kL¹(R²)= 0 and lim

t→∞k∇c(·,·+t)− ∇c_∞kL²(R²)= 0 means ”intermediate asymptotics” in original variables:

The stationary solution in self-similar variables

n_∞=M e^{c∞−|x|2/2} R

R²e^{c∞−|x|2/2}dx =−∆c_∞, c_∞=− 1

2πlog| · | ∗n_∞ Radial symmetry [ Naito ]

Uniqueness [ Biler, Karch, Lauren¸cot, Nadzieja ]

As |x| →+∞,n_∞is dominated bye^{−(1−ǫ)|x|2/2}for anyǫ∈(0,1) [ Blanchet, JD, Perthame ]

Bifurcation diagram ofkn_∞kL^∞(R²) as a function ofM:

M→lim0+kn_∞kL^∞(R²)= 0 [ Joseph, Lundgreen ] [ JD, Sta´nczy ]

The free energy in self-similar variables

∂n

∂t =∇h

n(logn−x+∇c)i F[n] :=

Z

R²

nlogn dx+ Z

R²

1

2|x|²n dx−1 2

Z

R²

n c dx satisfies

d

dtF[n(t,·)] =− Z

R²

n|∇(logn) +x− ∇c|²dx A last remark on8πand scalings: n^λ(x) =λ²n(λx) F[n^λ] =F[n]+

Z

R²

nlog(λ²)dx+

Z

R² λ⁻²−1

2 |x|²n dx+ 1 4π

Z

R²×R²

n(x)n(y) log1 λ dx dy F[n^λ]−F[n] =

2M−M² 4π

logλ+λ⁻²−1 2

Z

R²

|x|²n dx

First result on rates

Theorem

There exists a positive constant M^∗ such that, for any initial data n0∈L²(n⁻_∞¹dx)of mass M <M^∗ satisfying the above assumptions, there is a unique solution n∈C⁰(R⁺,L¹(R²))∩L^∞((τ,∞)×R²)for any τ >0

Moreover, there are two positive constants, C andδ, such that Z

R²|n(t,x)−n_∞(x)|² dx

n_∞ ≤C e^−δt ∀t >0 As a function of M, δis such thatlimM→0+δ(M) = 1

The condition M≤8πis necessary and sufficient for the global existence of the solutions, but there are two extra smallness conditions in our proof:

Uniform estimate: the method of the trap Spectral gap of a linearised operatorL

Proof of the first result on rates

First step: the trap

Second step: weightedH¹estimates Third step: linearization and spectral gap Fourth step: collecting the estimates

The parabolic-elliptic Keller and Segel system

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











∂u

∂t = ∆u− ∇ ·(u∇v) x ∈R², t >0

−∆v =u x ∈R², t >0 u(·,t = 0) =n0≥0 x ∈R² Initial conditions

n0∈L¹₊(R²,(1+|x|²)dx), n0logn0∈L¹(R²,dx), M:=

Z

R²

n0(x)dx <8π

First step: the trap

Decay Estimates of u ( t ) in L

( R

)

Lemma

For any M <M1, there exists C =C(M)such that, for any solution u∈C⁰(R⁺,L¹(R²))∩L^∞(R⁺

loc×R²)

ku(t)kL^∞(R²)≤C t⁻¹ ∀t>0 Themethod of the trap... prove that

H(ψ(t),M)≤0 where ψ(t) :=tku(·,t)k_L^∞(R²)

where z7→H(z,M)is a continuous function which is - negative on[0,z1)

- positive on (z1,z2)for somez1,z2such that0<z1<z2<∞ ψ is continuous andψ(0) = 0 =⇒ψ(t)≤z1≤z0(M)for anyt ≥0if H(z0(M),M) = sup_z∈[z1,z2]H(z,M)≥0

0 z H(z,M) H(z0(M),M)

H(z,M0(p)) z₀(M) z₀(M0(p))

−^M0(p)

2π

−_2π^M

Themethod of the trap amounts to prove thatH(z,M)≤0 implies that z=ψ(t)is bounded byz0(M)as long as H(z0(M),M)>0

Duhamel’s formula:

u(x,t0+t)− Z

R²

N(x−y,t)u(y,t0)dy

= Z t

0

Z

R²

N(x−y,t−s)∇ ·[u(y,t0+s)∇v(y,t0+s)] dy ds where N(x,t) =_4πt¹ e^−|x|2/(4t). Letκσ=k∂N/∂xi(·,1)kL^σ(R²)

ku(·,t0+t)kL^∞(R²)− 1

4πt ku(·,t0)kL¹(R²)

≤ X

i=1,2

Z t 0

∂N

∂xi

(·,t−s)∗h u ∂v

∂xi

(·,t0+s)i

L^∞(R²)

ds

≤ X

i=1,2

κσ

Z t 0

(t−s)^{−(1−σ1)−12}

u ∂v

∂xi

(·,t0+s)

L^ρ(R²)

ds

HLS inequality + H¨older and taket0=t 2t ku(·,2t)kL^∞(R²)− M

2π

≤2κσCHLS

π M^1p+1rt Z t

0

(t−s)^σ1−32(t+s)^1p+1r−2h

ψ(t)i2−p¹−¹r

ds with ψ(t) := sup_0≤s≤t2s ku(·,2s)k_L^∞(R²) and

t Z t

0

(t−s)^σ1−32(t+s)^1p+1r−2ds = σ 2−σ ψ(t)≤ M

2π+C0 ψ(t)θ

with C0=2κσCHLS

π M^1p+1r σ

2−σ , θ= 2−1 p−1

r Choice: H(z,M) =z−C0z^θ−M/(2π)

How small is the mass ?

The exponentsσ, ρ, p,q andr are related by

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

1

σ +¹_ρ = 1, 1< σ <2

1

p +¹_q =¹_ρ , p, q>2

1

r −¹_q = ¹₂, r >1 For the choice r= 4/3,q= 4, CHLS= 2√

π C0=^4κ√π^σM^p1+14 σ

2−σ withσ= _3p^4p₋₄

... there existsM0(p)such thatH(z0(M),M)>0 if and only if M <M0(p)andsup_p∈(4,+∞)M0(p) = limp→+∞M0(p)≈0.822663

<8π≈25.1327

Other norms: interpolation

Corollary

For any mass M<M1and all p∈[1,∞], there exists a positive constant C =C(p,M)withlim_M→0+C(p,M) = 0, such that

ku(t)k_L^p(R²)≤C t^−(1−p1) ∀t>0

RemarkThe above rates are optimal as can easily be checked using the self-similar solutions(n_∞, c_∞)

Second step: weighted H

estimates

L

and H

estimates in the self-similar variables

Consider the solution of

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∂n

∂t = ∆n− ∇ ·(n(∇c−x)) x∈R², t >0 c=− 1

2π log| · | ∗n x∈R², t >0 n(·,t = 0) =n0≥0 x∈R² For any p∈(1,∞]

kn(t)kL^p(R²)≤C1 ∀t >0 for some positive constantC1, and forp>2

2πk∇c(t)kL^∞ ≤ sup

x∈R²

Z

|x−y|≥1

n(t,y)

|x−y| dy

| {z }

+ sup

x∈R²

Z

|x−y|≤1

n(t,y)

|x−y| dy

| {z }

kn(t)kL^p(R²)≤C1 and k∇c(t)kL^∞(R²)≤C2 ∀t >0

Lemma

The constants C1and C2depend on M and are such that

Mlim→0+

Ci(M) = 0 i= 1, 2

Exponential weights

WithK =K(x) =e^|x|2/2, let us rewrite the equation fornas

∂n

∂t − 1

K ∇ ·(K∇n) =−∇c· ∇n+ 2n+n² Proposition

For any mass M ∈(0,M1), there is a positive constant C such that kn(t)kH¹(K)≤C ∀t >0

First ingredient[ M. Escobedo and O. Kavian ]: for anyq>2and ε >0, there exists a positive constantC(ε,q)such that

Z

n²K dx≤ε Z

|∇n|²K dx+C(ε,q)knk²L^q(R²)

L

( K ) estimate

Multiply the equation byn K and integrate by parts 1

2 d dt

Z

R²

|n|²K dx+ Z

R²

|∇n|²K dx

=− Z

R²

n∇c· ∇n K dx+ 2 Z

R²

n²K dx+ Z

R²

n³K dx

≤ε Z

R²

|∇n|²K dx+C and so

1 2

d dt

Z

R²

|n|²K dx+ (1−ε) Z

R²

|∇n|²K dx

| {z }

≥R

R2|n|²K dx

≤C

(expand the square inR

R²|∇(n K)|²K⁻¹dx≥0)

H

( K ) estimate (1/2)

LetS(t)be the semi-group generated by −K⁻¹∇ ·(K ∇·)onL²(K) n(t,x) =S(t)n0(x)−

Z t 0

S(t−s) (∇c·∇n)(s)ds+ Z t

0

S(t−s) (2n+n²)(s)ds

kn(t)kH¹(K)− kS(t)n0kH¹(K)

≤ Z t

0 kS(t−s) (∇c·∇n)(s)kH¹(K)ds+ Z t

0 kS(t−s) (2n+n²)(s)kH¹(K)ds Second ingredient: kS(t)hkH¹(K)≤κ(1 +t^−1/2)khkL²(K)

1

κ kn(t)kH¹(K)− kS(t)n0kH¹(K)

≤C2

Z t 1 +√¹

t−s

k∇n(s)kL²(K)ds+ (2 +C1) Z t

1 + √¹ t−s

kn(s)kL²(K)

H

( K ) estimate (2/2)

1

κkn(t+τ)kH¹(K)≤ 1 +√¹

t

C1+C3

Z t 0

1 + √¹ t−s

kn(s+τ)kH¹(K)ds

LetH(T) = sup_t∈(0,T)Rt 0

1 +^√_t¹_−s

kn(s+τ)kH¹(K)ds and choose T >0such that _2κ¹ =C3

RT 0

1 + ^√_T¹

−s

ds =C3 T + 2√ T 1

κH(T)≤ π+ 4√

T+T C1+ 1

2κH(T) =⇒ H(T)≤2 π+ 4√

T +T κC For any t∈(0,T)

1

κkn(t+τ)kH¹(K)≤ 1 +√¹

t

C1+C3H(T)≤ 1 +√¹

t

C1+2 π+ 4√

T +T

Conclusion: bound in H

( K )

The estimate 1

κkn(t+τ)kH¹(K)≤ 1 +√¹

t

C1+C3H(T)≤ 1 +√¹

t

C1+2 π+ 4√

T +T for any t∈(0,T)gives a bound onkn(T +τ)kH¹(K) for anyτ >0

Lemma

kn(t)k_H¹(K)≤C maxn 1,^√√^T

t

o ∀t >0

Actuallyn(t)can be bounded also inH¹(n⁻_∞¹)but further estimates are needed...

Third step: linearization and spectral gap

A linearized operator

Introducef andg defined by

n(x,t) =n_∞(x)(1 +f(x,t)) and c(x,t) =c_∞(x)(1 +g(x,t)) (f,g)is solution of the non-linear problem

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∂f

∂t − L(t,x,f,g) =− 1

n_∞∇ ·[f n_∞∇(g c_∞)] x∈R², t >0

−∆(c_∞g) =f n_∞ x∈R², t >0 where Lis the linear operator given by

L(t,x,f,g) = 1

n_∞∇ ·[n_∞∇(f −g c_∞)]

The conservation of mass is replaced here byR

f n_∞dx= 0

A spectral gap estimate

Proposition

For any M ∈(0,M2), for any f ∈H¹(n_∞dx)such that Z

R²

f n_∞dx = 0 =⇒ Z

R²

|∇f|²n_∞dx≥Λ(M) Z

R²

|f|²n_∞dx for someΛ(M)>0andlimM→0+Λ(M) = 1

Leth=√n_∞f =√

λe^{−|x|2/4+c∞/2}f λ|∇f|²n_∞=|∇h|²+|x|²

4 h²+1

4|∇c_∞|²h²+h∇h·(x−∇c_∞)−1

2x·∇c_∞h² (integrations by parts)

R

R²h∇h·x dx=−R

R²h²dx R

R²h∇h· ∇c_∞dx=¹₂R

R²h²(−∆c_∞)dx ≤¹₂ kn_∞kL^∞(R²)

R

R²h²dx

1 2

R

R²x· ∇c_∞h²dx≤ ^σ2σ⁻²¹

R

R²

|x|²

4 h²dx+¹_{4 σ}^σ2−²1

R

R²|∇c_∞|²h²dx

H

( n

) estimate

Assume thatn0/n_∞∈L²(n_∞)

There exists a constant C>0such that

|x|>1 =⇒

c_∞+M/(2π) log|x| ≤C n_∞K =e^c∞ behaves likeO(|x|^−M/(2π))as |x| → ∞

∂n

∂t −n_∞∇ · 1

n_∞∇n

= (∇c_∞− ∇c)· ∇n+ 2n+n²

Corollary

If M <M2, then any solution n is bounded in

L^∞(R⁺,L²(n⁻_∞¹dx))∩L^∞((τ,∞),H¹(n⁻_∞¹dx))

Fourth step: collecting the estimates

Proof of the exponential rate of convergence

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∂f

∂t − L(t,x,f,g) =− 1

n_∞∇ ·[f n_∞∇(g c_∞)] x∈R², t >0

−∆(c_∞g) =f n_∞ x∈R², t >0 Multiply by f n_∞and integrate by parts

1 2

d dt

Z

R²

|f|²n_∞dx+ Z

R²

|∇f|²n_∞dx

= Z

R²

∇f · ∇(g c_∞)n_∞dx

| {z }

=I

+ Z

R²

∇f · ∇(g c_∞)f n_∞dx

| {z }

=II

Cauchy-Schwarz’ inequality I=

Z

∇f · ∇(g c )n dx≤ k∇fk k∇(g c )k

first term

H¨older’s inequality (withq>2)

k∇(g c_∞)kL²(n∞dx)≤M^1/2−1/qkn_∞k^1/q_L^∞_(R²₎k∇(g c_∞)kL^q(R²)

HLS inequality (with 1/p= 1/2 + 1/q) k∇(g c_∞)k_L^q(R²)≤ 1

2π Z

R²

(f n_∞)∗_{| · |}¹

q

dx ¹_q

≤ CHLS

2π kf n_∞k_L^p(R²)

H¨older’s inequality: kf n_∞kL^p(R²)≤ kfkL²(n∞dx) kn_∞k^1/2_L^q/2_(R²₎ I=

Z

R²

∇f · ∇(g c_∞)f n_∞dx≤C_∗(M)kfkL²(n∞dx)k∇fkL²(n∞dx)

C_∗(M) :=CHLS(2π)⁻¹M^1/2−1/qkn_∞k^1/2_L^q/2_(R²₎ kn_∞k^1/q_L^∞(R²)→0as M→0

second term and conclusion

Useg c_∞=c−c_∞and the Cauchy-Schwarz inequality Z

R²

∇f· ∇(g c_∞)f n_∞dx≤ k∇c− ∇c_∞kL^∞(R²)

| {z }

≤2C2(M)ց0

kfkL²(n∞dx)k∇fkL²(n∞dx)

Spectral gap estimate pΛ(M)

| {z }

→1

kfkL²(n∞dx) ≤ k∇fkL²(n∞dx)

Withγ(M) := ^C∗(M)+2C√ ^2(M)

Λ(M) ց0,

1 2

d dt

Z

R²

|f|²n_∞dx≤ −[1−γ(M)]

Z

R²

|∇f|²n_∞dx

Uniqueness

Ifn1andn2are two solutions inC⁰(R⁺,L¹(R²))∩L^∞((τ,∞)×R²)for any τ >0, with f = (n2−n1)/n_∞ we also get

1 2

d dt

Z

R²|f|²n_∞dx≤ −[1−γ(M)] Λ(M) Z

R²|f|²n_∞dx As a consequence, if the initial condition is the same, thenn1=n2

B – Sobolev and

Hardy-Littlewood-Sobolev inequalities:

duality, flows

Gagliardo-Nirenberg inequalities

Consider the following sub-family of Gagliardo-Nirenberg inequalities kfk^L2p(R^d)≤Cp,dk∇fk^θL²(R^d)kfk¹L^−p+1θ(R^d)

withθ=θ(p) := ^p−_p¹_d+2−^d_p(d−2) 1<p≤ _d−^d2ifd ≥3 1<p<∞ifd= 2

[M. del Pino, J.D.]equality holds in iff =F_p with F_p(x) = (1 +|x|²)^−p−11 ∀x ∈R^d

and that all extremal functions are equal to Fp up to a multiplication by a constant, a translation and a scaling.

Ifd ≥3, the limit casep=d/(d−2)corresponds to Sobolev’s inequality [T. Aubin, G. Talenti]

Whenp→1, we recover the euclidean logarithmic Sobolev inequality in optimal scale invariant form[F. Weissler]

Ifd = 2andp→ ∞...

Onofri’s inequality as a limit case

Whend= 2, Onofri’s inequality can be seen as an endpoint case of the family of the Gagliardo-Nirenberg inequalities [J.D.]

Proposition

[J.D.]Assume that g ∈ D(R^d)is such thatR

R²g dµ= 0and let fp:=Fp(1 + g

2p)

Withµ(x) := _π¹(1 +|x|²)⁻², and dµ(x) :=µ(x)dx , we have

1≤ lim

p→∞Cp,2

k∇fk^θ(p)L²(R²)kfk¹L^−p+1θ(p)(R²)

kfk^L2p(R²)

= e^{161π RR2|∇g|2dx} R

R²e^g dµ

The standard form of the euclidean version of Onofri’s inequality is

Z Z 1 Z

Legendre duality: Onofri and log HLS

Legendre’s duality: F^∗[v] := sup R

R^du v dx−F[u]

F1[u] := log Z

R²

e^udµ

and F2[u] := 1 16π

Z

R²

|∇u|²dx+

Z

R²

uµdx Onofri’s inequality amounts toF1[u]≤F2[u]withdµ(x) :=µ(x)dx, µ(x) :=_π(1+¹_|x|2)²

Proposition

For any v ∈L¹₊(R²)withR

R²v dx = 1, such that v logv and (1 + log|x|²)v ∈L¹(R²), we have

F₁^∗[v]−F₂^∗[v] = Z

R²

v log v

µ

dx−4π Z

R²

(v−µ) (−∆)⁻¹(v−µ)dx≥0 [ Carlen-Loss, Beckner, Calvez-Corrias ]

A puzzling result of Carlen, Carrillo and Loss ( d ≥ 3)

[E. Carlen, J.A. Carrillo and M. Loss]The fast diffusion equation

∂v

∂t = ∆v^m t >0, x∈R^d with exponent m=d/(d+ 2), whend ≥3, is such that

Hd[v] :=

Z

R^d

v(−∆)⁻¹v dx−Sdkvk²

Ld+2^2d (R^d)

obeys to 1 2

d

dtHd[v(t,·)] = 1 2

d dt

Z

R^d

v(−∆)⁻¹v dx−Sdkvk²

Ld+2^2d (R^d)

=^d_(d^(d₋⁻₁₎²⁾2 Sdkuk^4/(d−L^q+1(R^1)d)k∇uk²L²(R^d)− kuk^2qL^2q(R^d)

withu=v^{(d−1)/(d+2)}andq=^d+1_d−1. If ^d_(d−^(d−₁₎²⁾2 Sd= (Cq,d)^2q, the r.h.s.

... and the two-dimensional case

Recall that (−∆)⁻¹v =Gd∗v with G_d(x) =_d−¹₂|S^d−1|⁻¹|x|^2−d ifd≥3 G2(x) =₂¹_π log|x|ifd = 2

Same computation in dimension d= 2withm= 1/2 gives kvk^L1(R²)

8 d dt

4π kvk^L1(R²)

Z

R²

v(−∆)⁻¹v dx− Z

R²

v logv dx

=kuk⁴L⁴(R²)k∇uk²L²(R²)−πkvk⁶L⁶(R²)

The r.h.s. is one of the Gagliardo-Nirenberg inequalities (d = 2, q= 3): π(C3,2)⁶= 1

The l.h.s. is bounded from below by the logarithmic

Hardy-Littlewood-Sobolev inequality and achieves its minimum if v =µwith

µ(x) := 1

π(1 +|x|²)² ∀x ∈R²

Sobolev and HLS

As it has been noticed by E. Lieb, Sobolev’s inequality in R^d,d ≥3, kuk²L^2∗(R^d) ≤Sdk∇uk²L²(R^d) ∀u∈ D^1,2(R^d) (1) and the Hardy-Littlewood-Sobolev inequality

Sdkvk²_Ld+2^2d

(R^d) ≥ Z

R^d

v(−∆)⁻¹v dx ∀v ∈L^d+22d (R^d) (2) are dualof each other. HereSd is the Aubin-Talenti constant and 2^∗= _d−^2d₂. Can we recover this using a nonlinear flow approach ? Can we improve it ?

Keller-Segel model: another motivation[Carrillo, Carlen and Loss]

and [Blanchet, Carlen and Carrillo]

Using a nonlinear flow to relate Sobolev and HLS

Consider the fast diffusionequation

∂v

∂t = ∆v^m t >0, x∈R^d (3)

If we define H(t) := Hd[v(t,·)], with Hd[v] :=

Z

R^d

v(−∆)⁻¹v dx−Sdkvk²

Ld+2^2d (R^d)

then we observe that 1

2H^′=− Z

R^d

v^m+1dx+ Sd

Z

R^d

v^d+22d dx _d²Z

R^d∇v^m· ∇v^d−2d+2 dx where v=v(t,·)is a solution of (??). With the choicem=^d_d+2⁻², we find that m+ 1 = _d+2^2d

A first statement

Proposition

[J.D.]Assume that d ≥3andm=^d−_d+2². If v is a solution of (??)with nonnegative initial datum in L^2d/(d+2)(R^d), then

1 2

d dt

Z

R^d

v(−∆)⁻¹v dx−Sdkvk²

Ld+2^2d(R^d)

= Z

R^d

v^m+1dx _d²h

Sdk∇uk²L²(R^d)− kuk²L^2∗(R^d)

i≥0

The HLS inequality amounts to H≤0 and appears as a consequence of Sobolev, that is H^′ ≥0 if we show thatlim sup_t>0H(t) = 0

Notice that u=v^mis an optimal function for(??)ifv is optimal

Improved Sobolev inequality

By integrating along the flow defined by(??), we can actually obtain optimal integral remainder terms which improve on the usual Sobolev inequality (??), but only whend ≥5for integrability reasons

Theorem

[J.D.]Assume that d ≥5and let q=_d^d+2₋₂. There exists a positive constant C ≤ 1 + ²_d

1−e^−d/2

Sd such that Sdkw^qk²

Ld+2^2d(R^d)− Z

R^d

w^q(−∆)⁻¹w^qdx

≤ C kwk

8 d−2

L^2∗(R^d)

h

k∇wk²L²(R^d)−Sdkwk²L^2∗(R^d)

i

for any w∈ D^1,2(R^d)

Solutions with separation of variables

Consider the solution of ^∂v_∂t = ∆v^m vanishing att =T: vT(t,x) =c(T −t)^α(F(x))^d−2d+2 where F is the Aubin-Talenti solution of

−∆F =d(d−2)F^{(d+2)/(d−2)} Letkvk∗:= sup_x∈Rd(1 +|x|²)^d+2|v(x)|

Lemma

[M. delPino, M. Saez],[J. L. V´azquez, J. R. Esteban, A. Rodr´ıguez]

For any solution v with initial datum v0∈L^2d/(d+2)(R^d), v0>0, there exists T >0,λ >0and x0∈R^d such that

t→limT−

(T −t)^−1−m1 kv(t,·)/v(t,·)−1k∗= 0

Improved inequality: proof (1/2)

J(t) :=R

R^dv(t,x)^m+1dx satisfies

J^′ =−(m+ 1)k∇v^mk²L²(R^d)≤ −m+ 1 Sd

J^1−2d Ifd ≥5, then we also have

J^′′= 2m(m+ 1) Z

R^d

v^m−1(∆v^m)²dx≥0 Notice that

J^′

J ≤ −m+ 1 Sd

J^−2d ≤ −κ with κT = 2d d+ 2

T Sd

Z

R^d

v₀^m+1dx −²d

≤d 2

Improved inequality: proof (2/2)

By the Cauchy-Schwarz inequality, we have J^′2

(m+ 1)² =k∇v^mk⁴L²(R^d)= Z

R^d

v^(m−1)/2∆v^m·v^(m+1)/2dx 2

≤ Z

R^d

v^m−1(∆v^m)²dx Z

R^d

v^m+1dx=CstJ^′′J so thatQ(t) :=k∇v^m(t,·)k²L²(R^d)

R

R^dv^m+1(t,x)dx−(d−2)/d

is monotone decreasing, and

H^′= 2 J (SdQ−1), H^′′= J^′

J H^′+ 2 J SdQ^′ ≤J^′ J H^′ ≤0 H^′′≤ −κH^′ with κ= 2d

d+ 2 1 Sd

Z

R^d

v₀^m+1dx −2/d

′ −κT

d = 2: Onofri’s and log HLS inequalities

H2[v] :=

Z

R²

(v−µ) (−∆)⁻¹(v−µ)dx− 1 4π

Z

R²

v log v

µ

dx Withµ(x) := _π¹(1 +|x|²)⁻². Assume thatv is a positive solution of

∂v

∂t = ∆ log (v/µ) t >0, x∈R² Proposition

If v =µe^u/2is a solution with nonnegative initial datum v0in L¹(R²) such thatR

R²v0dx= 1, v0 logv0∈L¹(R²)and v0logµ∈L¹(R²), then d

dtH2[v(t,·)] = 1 16π

Z

R²

|∇u|²dx− Z

R²

e^u2 −1 u dµ

≥ 1 16π

Z

R²|∇u|²dx+ Z

R²

u dµ−log Z

R²

e^udµ

≥0

C: Keller-Segel model

1

Small mass results

2

Spectral analysis

3

Collecting estimates: towards exponential

convergence

Keller-Segel with subcritical mass in self-similar variables











∂n

∂t = ∆n− ∇ ·(n(∇c−x)) x∈R², t >0 c=− 1

2π log| · | ∗n x∈R², t >0 n(·,t = 0) =n0≥0 x∈R²

tlim→∞kn(·,·+t)−n_∞kL¹(R²)= 0 and lim

t→∞k∇c(·,·+t)− ∇c_∞kL²(R²)= 0 n_∞=M e^{c∞−|x|2/2}

R

R²e^{c∞−|x|2/2}dx =−∆c_∞, c_∞=− 1

2πlog| · | ∗n_∞

A parametrization of the solutions and the linearized operator

[ Campos, JD ]

−∆c=M e^−12|x|2+c R

R²e^−12|x|2+c dx Solve

−φ^′′−1

r φ^′=e^−12r2+φ, r >0 with initial conditionsφ(0) =a,φ^′(0) = 0and get

M(a) := 2π Z

R²

e^−12r2+φadx

na(x) =M(a) e^{−12r2+φa(r)} 2πR

R²r e^−12r2+φa dx =e^{−12r2+φa(r)} With−∆ϕf =naf, consider the operator defined by

Spectrum of L (lowest eigenvalues only)

5 10 15 20 25

1 2 3 4 5 6 7

Figure: The lowest eigenvalues of−L(shown as a function of the mass) are 0, 1 and 2, thus establishing that the spectral gap of−Lis 1

Simple eigenfunctions

Kernel Letf0=_∂M^∂ c_∞ be the solution of

−∆f0=n_∞f0

and observe thatg0=f0/c_∞is such that 1

n_∞∇ · n_∞∇(f0−c_∞g0)

=:Lf0= 0 Lowest non-zero eigenvalues f1:=_n¹_∞ ^∂n_∂x^∞₁ associated with g1= _c¹_∞ ^∂c_∂x^∞₁ is an eigenfunction ofL, such that −Lf1=f1

WithD:=x· ∇, letf2= 1 + ¹₂D logn_∞= 1 +_2n¹_∞D n_∞. Then

−∆ (D c_∞) + 2 ∆c_∞=D n_∞= 2 (f2−1)n_∞ and sog := ¹ (−∆)⁻¹(n f )is such that−Lf = 2f

Functional setting...

F[n] :=

Z

R²

nlog n

n_∞

dx−1 2

Z

R²

(n−n_∞) (c−c_∞)dx achieves its minimum for n=n_∞ according to log HLS and

Q1[f] = lim

ε→0

1

ε²F[n_∞(1 +εf)]≥0 ifR

R²f n_∞dx= 0. Notice thatf0generates the kernel ofQ1

Lemma

For any f ∈H¹(R²,n_∞dx)such thatR

R²f n_∞dx= 0, we have Z

R²

|∇(g c_∞)|²n_∞dx≤ Z

R²

|f|²n_∞dx

... and eigenvalues

Withg such that−∆(g c_∞) =f n_∞, Q1determines a scalar product hf1,f2i:=

Z

R²

f1f2n_∞dx− Z

R²

f1n_∞(g2c_∞)dx

on the orthogonal tof0 inL²(n_∞dx)and withG2(x) :=−2π¹ log|x| Q2[f] :=

Z

R²

|∇(f −g c_∞)|²n_∞dx with g = 1

c_∞G2∗(f n_∞) is a positive quadratic form, whose polar operator is the self-adjoint operatorL

hf,Lfi= Q2[f] ∀f ∈ D(L2) Lemma

In this setting,Lhas pure discrete spectrum and its lowest eigenvalue is

An interpolation inequality induced by log HLS

Lemma

For any f ∈L²(R²,n_∞dx)such thatR

R²f f0n_∞dx = 0holds, we have

− 1 2π

Z Z

R²×R²

f(x)n_∞(x) log|x−y|f(y)n_∞(y)dx dy

≤ Z

R²

|f|²n_∞dx where g c_∞=G2∗(f n_∞)and, ifR

R²f n_∞dx= 0holds, Z

R²

|∇(g c_∞)|²dx≤ Z

R²

|f|²n_∞dx

Equalities in the above inequality then holds if and only if f = 0

A new Onofri type inequality

The spectral gap inequality of Lis a refined version of Theorem (Onofri type inequality)

For any M ∈(0,8π), if n_∞=M ^ec∞−

12|x|2

R

R2e^{c∞ −12|x|2}dx with c_∞= (−∆)⁻¹n_∞, dµM= _M¹ n_∞dx , we have the inequality

log Z

R²

e^φdµM

− Z

R²

φdµM≤ 1 2M

Z

R²

|∇φ|²dx ∀φ∈ D0^1,2(R²)